☰ अधिक माहितीसाठी येथे क्लीक करा

Question 1:

Different mathematical operations are given in the two rows below. Find out the number you get in each operation and make equations.
16 ÷ 2,           5 × 2,            9 + 4,              72 ÷ 3,
4 + 5             8 × 3,          19 $-$ 10,           10 $-$ 2,
37 $-$ 27,         6 + 7

16 ÷ 2 = 8

5 × 2 = 10

9 + 4 = 13

72 ÷ 3 = 24

4 + 5 = 9

8 × 3 = 24

19 − 10 = 9

10 − 2 = 8

37 − 27 = 10

6 + 7 = 13

So,

16 ÷ 2 = 10 − 2

5 × 2 = 37 − 27

9 + 4 = 6 + 7

72 ÷ 3 = 8 × 3

4 + 5 = 19 − 10

Question 1:

Rewrite the following using a letter.
(1) The sum of a certain number and 3.
(2) The difference obtained by subtracting 11 from another number.
(3) The product of 15 and another number.
(4) Four times a number is 24.

(1)
Let the certain number be x.

∴ Sum of a certain number and 3 = x + 3

(2)
Let another number be x.

∴ Difference obtained by subtracting 11 from another number = x − 11

(3)
Let another number be x.

∴ Product of 15 and another number = 15 × x = 15x

(4)
Let the number be x.

Four time a number = 24

∴ 4 × x = 24

⇒ 4x = 24

Question 2:

Find out which operation must be done on both sides of these equations in order to solve them.
(1) + 9 = 11       (2) $-$ 4 = 9          (3) 8x = 24        (4)

(1) Subtract 9 from both sides

+ 9 = 11

⇒ + 9 − 9 = 11 − 9               (Subtract 9 from both sides)

⇒ x + 0 = 2

⇒ x = 2

(2) Add 4 to both sides

− 4 = 9

⇒ − 4 + 4 = 9 + 4         (Add 4 to both sides)

⇒ x + 0 = 13

⇒ x = 13

(3) Divide both sides by 8

8x = 24

⇒ $\frac{8x}{8}=\frac{24}{8}$          (Divide both sides by 8)

⇒ x = 3

(4) Multiply both sides by 6

⇒              (Multiply both sides by 6)

⇒ x = 18

Question 3:

Given below are some equations and the values of the variables. Are these values the solutions to those equations?

 No Equation Value of the variable Solution(Yes / No) 1 y − 3 = 11 y = 3 No 2 17 = n + 7 n = 10 3 30 = 5x x = 6 4 m2= 14 m = 7

 No Equation Value of the variable Solution(Yes / No) 1 y − 3 = 11 y = 3 No 2 17 = n + 7 n = 10 Yes 3 30 = 5x x = 6 Yes 4 m2= 14 m = 7 No

Explanation

(1)
When y = 3,

LHS = y − 3 = 3 − 3 = 0

RHS = 11

Since LHS ≠ RHS, so y = 3 is not a solution of equation y − 3 = 11.

(2)
When n = 10,

RHS = n + 7 = 10 + 7 = 17

LHS = 17

Since LHS = RHS, so n = 10 is a solution of equation 17 = n + 7.

(3)
When x = 6,

RHS = 5x = 5 × 6 = 30

LHS = 30

Since LHS = RHS, so x = 6 is a solution of equation 30 = 5x.

(4)
When m = 7,

LHS = $\frac{m}{2}=\frac{7}{2}$

RHS = 14

Since LHS ≠ RHS, so m = 7 is not a solution of equation .

Question 4:

Solve the following equations.
(1)  = 1    (2) 8 = t + 5       (3) 4x = 52                  (4) 19 = 4
(5)        (6) x + 10 = 5     (7)       (8) p + 4 = $-$ 1

(1)
y − 5 = 1

⇒ y − 5 + 5 = 1 + 5         (Add 5 to both sides)

⇒ y + 0 = 6

⇒ y = 6

Thus, the solution of the given equation is y = 6.

(2)
8 = t + 5

⇒ 8 − 5 = t + 5 − 5         (Subtract 5 from both sides)

⇒ 3 = t + 0

⇒ 3 = t

Thus, the solution of the given equation is t = 3.

(3)
4x = 52

⇒ $\frac{4x}{4}=\frac{52}{4}$            (Divide both sides by 4)

⇒ x = 13

Thus, the solution of the given equation is x = 13.

(4)
19 = m − 4

⇒ 19 + 4 = m − 4 + 4         (Add 4 to both sides)

⇒ 23 = m + 0

⇒ 23 = m

Thus, the solution of the given equation is m = 23.

(5)
$\frac{p}{4}=9$

⇒ $\frac{p}{4}×4=9×4$        (Multiply both sides by 4)

⇒ p = 36

Thus, the solution of the given equation is p = 36.

(6)
x + 10 = 5

⇒ x + 10 − 10 = 5 − 10            (Subtract 10 from both sides)

⇒ x + 0 = −5

⇒ = −5

Thus, the solution of the given equation is = −5.

(7)
m − 5 = −12

⇒ m − 5 + 5 = −12 + 5           (Add 5 to both sides)

⇒ m + 0 = −7

⇒ m = −7

Thus, the solution of the given equation is = −7.

(8)
p + 4 = −1

⇒ p + 4 − 4 = −1 − 4           (Subtract 4 from both sides)

⇒ p + 0 = −5

⇒ p = −5

Thus, the solution of the given equation is = −5.

Question 5:

Write the given information as an equation and find its solution.
(1) Haraba owns some sheep. After selling 34 of them in the market, he still has 176 sheep. How many sheep did Haraba have at first?

(2) Sakshi prepared some jam at home and filled it in bottles. After giving away 7 of the bottles to her friends, she still has 12 for herself. How many bottles had she made in all? If she filled 250g of jam in each bottle, what was the total weight of the jam she made?

(3) Archana bought some kilograms of wheat. She requires 12kg per month and she got enough wheat milled for 3 months. After that, she had 14 kg left. How much wheat had Archana bought altogether?

(1)
Let the number of sheeps with Haraba at first be x.

According to the given condition,

Number of sheeps with Haraba at first − Number of sheeps sold in the market = Number of sheeps left with Haraba

∴ x − 34 = 176

⇒ x − 34 + 34 = 176 + 34          (Add 34 to both sides)

⇒ x + 0 = 210

⇒ x = 210

Thus, there were 210 sheeps with Haraba at first.

(2)
Let the total number of jam bottles made by Sakshi be x.

According to the given condition,

Total number of jam bottles made by Sakshi − Number of jam bottles given to her friends = Number of jam bottles left with Sakshi

∴ x − 7 = 12

⇒ x − 7 + 7 = 12 + 7          (Add 7 to both sides)

⇒ x + 0 = 19

⇒ x = 19

So, the total number of jam bottles made by Sakshi are 19.

Weight of jam in each bottle = 250 g

∴ Total weight of the jam

= Weight of jam in each bottle × Number of bottles

= 250 g × 19

= 4750 g

= 4.75 kg            (1 kg = 1000 g)

Thus, the total weight of the jam made by Sakshi is 4750 g or 4.75 kg.

(3)
Let the weight of wheat bought by Archana altogether be x kg.

According to the given condition,

Weight of wheat bought by Archana altogether − Weight of the wheat used in 3 months = Amount of wheat left with Archana

∴ kg − 12 kg/month × 3 months = 14 kg

⇒ x − 36 = 14

⇒ x − 36 + 36 = 14 + 36          (Add 36 to both sides)

⇒ x + 0 = 50

⇒ x = 50

Thus, the weight of wheat bought by Archana altogether is 50 kg.
Std 6 th Mathematics- 10. Equations Reviewed by Amol Uge on January 09, 2019 Rating: 5